The present invention relates to a tomograph which makes use of the nuclear magnetic resonance (which will be abbreviated to "NMR") and which is used for medical diagnosis.
The problem of a timing error in the measurement of an NMR signal is solved in the prior art by displaying the absolute (spectral) value of an image obtained (as should be referred to the third line from the bottom of pp. 232, "Technical alternatives in nuclear magnetic resonance (NMR) imaging", SPIE 1983, written by P. A. Bottomley, for example).
This problem will be discussed in detail in the following.
Now, for simplicity of discussion, consideration will be taken into a one-dimensional signal, but a similar discussion applies to a two-dimensional signal. If the NMR signal is expressed by f(t), the NMR signal has its real part obtained in the form of an even function and its imaginary part obtained in the form of an odd function according to its characteristics, as shown in (i) and (ii) of FIG. 1(a).
Hence, the Fourier transformation F(x) of f(t) is expressed by a real function in accordance with its characteristics: EQU F(x)=F[f(t)] (1).
If F(x) is expressed in a polar form, therefore, the phase component is zero: EQU F(x)=A(x)e.sup.jo =A(x) (2).
Here, if the measuring timing of f(t) deviates by a, as shown in FIG. 1(b), then: EQU f'(t)=f(t+a) (3).
The Fourier-transformed component F'(x) is determined by: ##EQU1## In other words, the phase component .theta., which should not intrinsically exist, as shown in (i) of FIG. 1(c), appears in the form proportional to x, as shown in (ii) of FIG. 1(c).
If a phase is to be determined from an actual data, not only the phase deviation ax due to the timing deviation a but also the phase deviation due to the characteristics of a receiver and the phase deviation due to inhomogeneities of the magnetic field are superposed, the phase determined is not correct in a low-level position of the measured signal but takes a random value.
In the NMR imaging, by noting that F(x) gives the value sought for, the image is generally displayed in an absolute value by making use of the characteristics of the Fourier transformation that the absolute value of F(x) is not influenced by the time deviation of f(t): EQU .vertline.=F(x).vertline.=.vertline.A(x)e.sup.jax .vertline.=.vertline.A(x).vertline.-F(x) (5)
The method described above is effective because of its simple processing but is accompanied by a defect that a correct image cannot be obtained unless the value of F(x) is positive, because the absolute value is taken.